Answer:
Step-by-step explanation:
H₀ : The distribution of the sample agrees with the population distribution.
H₁ : the distribution of the does not agree with the population distribution.
Using the Chi-square test statistics
we will compute the expected frequencies
Now expected frequency for Washington will be
Fe = Npi = 1000×51% = 5100
expected frequency for Oregon
Fe = Npi = 1000×30% = 300
expected frequency for Idaho
Fe = Npi = 1000×11% = 110
expected frequency for Montana
Fe = Npi = 1000×8% = 80
The Chi=square test statistics is;
x² = ∑ [ (Fo - Fe)² / Fe )
now we substitute
Given our Fo ( 450 in Washington, 340 in Oregon, 150 in Idaho, and 60 in Montana) from the question
x² = (450-510)²/510 + (340-300)²/300 + (150-110)²/110 + (60-80)²/80
x²= 31.9376
so the chi-square test statistics is 31.9376
Now the chi-square critical value
first we compute for degree of freedom
d.f = k -1 = 4 - 1 = 3
So from the critical value table, degree of freedom 3 and significance level 0.05,
the chi-square critical value is 7.8147
Therefore chi-square test statistics 31.9376 is greater than the chi-square critical value 7.8147, so NULL HYPOTHESIS IS REJECTED at 5% significance.
There is sufficient evidence to warrant rejection of the claim that the distribution of the sample agrees with the distribution of the state populations.
